R 搜索不在列表中的下一个最近的元素

我试图从26个字母的向量中替换2个字母（重复）

我的表中已经有26个字母中的13个（键），所以替换字母不应该在这13个“键”中

我正在尝试编写代码，用下一个字母表取代

C&S

，而下一个字母表不应该是“键”的一部分

下面的代码将repeat

C替换为D

，

S替换为T

，但这两个字母都在我的“键”中。有人知道我如何实现这个条件，以便在“key”中已经存在要替换的字母时代码将重新运行循环吗

# alphabets <- toupper(letters)
keys <- c("I", "C", "P", "X", "H", "J", "S", "E", "T", "D", "A", "R", "L")
repeats <- c("C", "S")
index_of_repeat_in_26 <- which(repeats %in% alphabets)
# index_of_repeat_in_26 is 3 , 19
# available_keys <- setdiff(alphabets,keys)
available <- alphabets[available_keys]
# available <-  c("B", "F", "G", "K", "O", "Q", "U", "V", "W", "Y", "Z")
index_available_keys <- which(alphabets %in% available_keys)
# 2  6  7 11 15 17 21 22 23 25 26

for (i in 1:length(repeat)){
    for(j in 1:(26-sort(index_of_repeat_in_26)[1])){
        if(index_of_repeat_in_26[i]+j %in% index_available_keys){
            char_to_replace_in_key[i] <- alphabets[index_of_capital_repeat_in_26[i]+1]
        }
        else{
            cat("\n keys not available to replace \n")
        }
    }
}

---

#alphabets
键检查
while
函数。u相关，
字母
产生与
toupper（字母）
相同的输出。我尝试了while循环，但程序仍在运行，而（char_to_replace_in_key！=（可用））{for（I in 1:length（repeat））{for（j in 1:（26排序（索引_repeat_in_26）[1]））{char_to_replace_in_key[t]Joel，谢谢，非常有用。是的，对于这个查询，它工作得非常出色。当我有25个键（A-Y），4个重复（C，S，M，N），只有一个字母表要替换（Z）。上面的代码是用Z替换所有重复。我想要的是它应该只用Z替换C，而对于剩余的S，M，N，它应该显示“key not available”消息。你知道如何通过if语句或类似的语句通过for循环实现这个条件吗？@mr_swap ohhh！我不知道这个条件。好吧应该做了！现在我没有访问RStudio的权限。以后再更新吗？问题仍然存在。现在我给了24个键（A-X），4个重复（CSMN）。它仍然用Y替换C，保持其他键不变。它工作：-），你能解释一下函数是如何逐行执行此操作的吗？
keys <- c("I", "C", "P", "X", "H", "J", "S", "E", "T", "D", "A", "R", "L")
repeats <- c("C", "S")

y = sort(setdiff(LETTERS, keys))       # get the letters not present in 'keys' 
y = factor(y, levels = LETTERS)        # make them factor so that we can do numeric comparisons with the levels
y1 = as.numeric(y)                     # keep them numeric to compare
z  = factor(repeats, levels = LETTERS)
z1 = as.numeric(z)

func <- function(x) {         # so here, in each iteration, the index(in this case 1:4 gets passed)
           xx = y1 - z1[x]    # taking the difference between each 'repeat' element from all 'non-keys'
           xx = which(xx>0)[1]# choose the one with smallest difference(because 'y1'  is already sorted. So the first nearest non-key gets selected
           r = y[xx]          # extract the corresponding 'non-key' element            
           y <<- y[-xx]       # after i get the closest letter, I remove that from global list so that it doesn't get captured the next time
           y1 <<- y1[-xx]     # similarily removed from the equivalent numeric list
           r                  # return the extracted 'closest non-key' chracter 
 }

 # sapply is also a for-loop by itself, in which a single element get passed ro func at a time. 
 # Here 'seq_along' is used to pass the index. i.e. for 'C' - 1, for 'S' - 2 , etc gets passed.

 ans = sapply(seq_along(repeats), func)

if (any(is.na(ans))){
  cat("\n",paste0("keys not available to replace for ",
           paste0(repeats[which(is.na(ans))], collapse = ",")) ,
      "\n")
  ans <- ans[!is.na(ans)]
}

# example 2 with :
repeats <- c("Y", "Z")
# output : 
# keys not available to replace for Z 
# ans
# [1] Z